Introduction to Abstract Algebra, Solutions Manual, 4th Edition

8.4 The Sylow Theorems

1. Since |S₄| = 2³ · 3, the Sylow 3-subgroups are all cyclic of order 3, and thus have the form P = γ , γ = (i j k). Now σ(123)σ⁻¹ = (σ(1)σ(2) σ(3)) for all σ S₄ (Lemma 3 §2.8) so let where {1, 2, 3, 4} = {i, j, k, x}. Then σ(123)σ⁻¹ = (ijk) so γ and (1 2 3) are conjugate. Hence P and (123) are conjugate.

3. P is a Sylow p-subgroup of N(P), being a p-subgroup of maximal order. It is unique because it is normal in N(P).

5. Let |G| = 1001 = 7 · 11 · 13. We have n₇ = 1, 11, 13, 143 and n₇ ≡ 1 (mod 7), so n₇ = 1. Similarly n₁₁ = 1, 7, 13, 91 and n₁₁ ≡ 1 (mod 11), so n₁₁ = 1; and n₁₃ = 1, 7, 11, 77 and n₁₃ ≡ 1 (mod 13) so n₁₃ = 1. Thus let H G, K G, L G have order | ...